You should talk about how it evaporates then it turn into a gas then back in to a liquid
Answer: m = 11.2 g C7H16
Explanation: First convert the mass of CO2 to moles. Then do the mole ratio between CO2 and C7H16 which is 7:1. Finally convert the moles of C7H16 to the mass of C7H16.
Solution attached.
The answer for the following problem is mentioned below.
- <u><em>Therefore the final moles of the gas is 14.2 × </em></u>
<u><em> moles.</em></u>
Explanation:
Given:
Initial volume (
) = 230 ml
Final volume (
) = 860 ml
Initial moles (
) = 3.8 × moles
To find:
Final moles (
)
We know;
According to the ideal gas equation;
P × V = n × R × T
where;
P represents the pressure of the gas
V represents the volume of the gas
n represents the no of the moles of the gas
R represents the universal gas constant
T represents the temperature of the gas
So;
V ∝ n
= 
where,
() represents the initial volume of the gas
() represents the final volume of the gas
() represents the initial moles of the gas
() represents the final moles of the gas
Substituting the above values;
= 
= 14.2 × moles
<u><em>Therefore the final moles of the gas is 14.2 × </em></u><u><em> moles.</em></u>
Answer:
0.7561 g.
Explanation:
- The hydrogen than can be prepared from Al according to the balanced equation:
<em>2Al + 6HCl → 2AlCl₃ + 3H₂,</em>
It is clear that 2.0 moles of Al react with 6.0 mole of HCl to produce 2.0 moles of AlCl₃ and 3.0 mole of H₂.
- Firstly, we need to calculate the no. of moles of (6.8 g) of Al:
no. of moles of Al = mass/atomic mass = (6.8 g)/(26.98 g/mol) = 0.252 mol.
<em>Using cross multiplication:</em>
2.0 mol of Al produce → 3.0 mol of H₂, from stichiometry.
0.252 mol of Al need to react → ??? mol of H₂.
∴ the no. of moles of H₂ that can be prepared from 6.80 g of aluminum = (3.0 mol)(0.252 mol)/(2.0 mol) = 0.3781 mol.
- Now, we can get the mass of H₂ that can be prepared from 6.80 g of aluminum:
mass of H₂ = (no. of moles)(molar mass) = (0.3781 mol)(2.0 g/mol) = 0.7561 g.
Oxygen is an element, symbol O
Magnesium oxide is an ionic compound, formula MgO
